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Question 4.1: The function f is periodic with period T=4, defined as: f(x) = x^2 - 4 & if 0 x < 2 \\ b - ax & if 2 x < 4 It is given that f is continuous at x=2. For a periodic function to be continuous over its entire domain, it must also satisfy f(0) = _x 4^- f(x). a) The value of the constant b. Step 1: Use the continuity at x=2. For f(x) to be continuous at x=2, the left-hand limit must equal the right-hand limit: _x 2^- f(x) = _x 2^+ f(x) Using the first piece for x < 2: _x 2^- (x^2 - 4) = 2^2 - 4 = 4 - 4 = 0 Using the second piece for x 2: _x 2^+ (b - ax) = b - a(2) = b - 2a Equating these, we get our first equation: b - 2a = 0 b = 2a (*) Step 2: Use the continuity condition for the periodic function at the interval boundaries. For a periodic function defined on [0, T) to be continuous, f(0) must equal _x T^- f(x). Here T=4. f(0) = 0^2 - 4 = -4 _x 4^- f(x) = _x 4^- (b - ax) = b - a(4) = b - 4a Equating these, we get our second equation: b - 4a = -4 (**) Step 3: Solve the system of equations for a and b. Substitute b = 2a from () into (*): 2a - 4a = -4 -2a = -4 a = 2 Now substitute a=2 back into b=2a: b = 2(2) = 4 The value of the constant b is 4. b) f(13) and f(-1.1). Step 1: Calculate f(13) using the periodicity. Since the period is T=4, f(x+4k) = f(x) for any integer k. We find the equivalent value in the interval [0, 4): 13 = 3 × 4 + 1 So, f(13) = f(1). Since 0 1 < 2, we use f(x) = x^2 - 4: f(1) = 1^2 - 4 = 1 - 4 = -3 Therefore, f(13) = -3. Step 2: Calculate f(-1.1) using the periodicity. We find the equivalent value in the interval [0, 4): -1.1 + 1 × 4 = 2.9 So, f(-1.1) = f(2.9). Since 2 2.9 < 4, we use f(x) = b - ax. With a=2 and b=4, this is f(x) = 4 - 2x: f(2.9) = 4 - 2(2.9) = 4 - 5.8 = -1.8 Therefore, f(-1.1) = -1.8. c) Investigate the parity of f in the interval 0 x < 2. Parity (even or odd) is typically defined for functions whose domain is symmetric about the origin (e.g., [-L, L]). The interval [0, 2) is not symmetric. However, if we consider the expression f(x) = x^2 - 4 for x [0, 2), this expression itself represents an even function because g(-x) = (-x)^2 - 4 = x^2 - 4 = g(x). But the function f(x) as a whole, considering its periodic nature, is neither even nor odd. For example, f(1) = -3. If f were even, f(-1) would be -3. If f were odd, f(-1) would be 3. Using periodicity, f(-1) = f(-1+4) = f(3). From part (a), f(x) = 4-2x for 2 x < 4. So, f(3) = 4 - 2(3) = 4 - 6 = -2. Since f(-1) = -2, and f(1) = -3: f(-1) f(1) (so not even). f(-1) -f(1) (since -(-3) = 3, so not odd). Therefore, the function f(x) is neither even nor odd. d) Sketch the graph of f(x) in the interval -4 x < 8. The function is f(x) = x^2 - 4 & if 0 x < 2 \\ 4 - 2x & if 2 x < 4 with period 4. We need to sketch it over three periods: [-4, 0), [0, 4), and [4, 8). For the interval [0, 4): • For 0 x < 2, f(x) = x^2 - 4. This is a parabola. f(0) = -4 f(1) = -3 f(2) = 0 (approaching from left) • For 2 x < 4, f(x) = 4 - 2x. This is a straight line. f(2) = 0 f(3) = -2 f(4) = -4 (approaching from left, f(4) is technically f(0) due to periodicity) For the interval [4, 8): Using f(x) = f(x-4): • For 4 x < 6 (i.e., 0 x-4 < 2), f(x) = (x-4)^2 - 4. f(4) = -4 f(5) = -3 f(6) = 0 • For 6 x < 8 (i.e., 2 x-4 < 4), f(x) = 4 - 2(x-4). f(6) = 0 f(7) = -2 f(8) = -4 (approaching from left) For the interval [-4, 0): Using f(x) = f(x+4): • For -4 x < -2 (i.e., 0 x+4 < 2), f(x) = (x+4)^2 - 4. f(-4) = -4 f(-3) = -3 f(-2) = 0 • For -2 x < 0 (i.e., 2 x+4 < 4), f(x) = 4 - 2(x+4). f(-2) = 0 f(-1) = -2 f(0) = -4 (approaching from left) The graph will consist of three identical segments. Each segment starts at y=-4, curves up to y=0, then goes down linearly to y=-4. (A detailed sketch cannot be provided in this text format, but the description above outlines the shape and key points for plotting.) Question ii) Prove by contradiction that (1)/(sqrt(2)) is irrational. Step 1: Assume the opposite (for contradiction). Assume that (1)/(sqrt(2)) is a rational number. If (1)/(sqrt(2)) is rational, then it can be expressed as a fraction (p)/(q), where p and q are integers, q 0, and p and q have no common factors (i.e., the fraction is in its simplest form). (1)/(sqrt(2)) = (p)/(q) Step 2: Manipulate the equation to isolate sqrt(2). Cross-multiply the equation: q = psqrt(2) Divide both sides by p (since p must be non-zero, otherwise (1)/(sqrt(2))=0, which is false): sqrt(2) = (q)/(p) Step 3: Analyze the implication. Since p and q are integers and p 0, the fraction (q)/(p) represents a rational number. This implies that sqrt(2) is a rational number. Step 4: State the contradiction. However, it is a well-established mathematical fact that sqrt(2) is an irrational number. The conclusion that sqrt(2) is rational directly contradicts this known fact. Step 5: Conclude. Since our initial assumption (that (1)/(sqrt(2)) is rational) leads to a contradiction, the assumption must be false. Therefore, (1)/(sqrt(2)) is an irrational number. What's next?